Applications of Integration

$\frac{12}{4}$

$\frac{8}{3}$

$\sqrt[3]{4}$

$\frac{4}{3}$

The integral of the function’s derivative over $(a, b)$ must equal zero.

There must exist points where the function attains its maximum and minimum on $(a, b)$.

The derivative of $f(x)$ must exist at every point in $(a, b)$.

The function $f(x)$ must be continuous on the closed interval $[a, b]$.

$\sqrt{\left(\int_2^6 [g(t)]^2 , dt\right)}$

$\frac{g(6) - g(2)}{4}$

$\frac{1}{4} \int_2^6 g(t) , dt$

$\int_2^6 g(t) , dt$

$\frac{1}{2}$

$\frac{2}{3}$

$\frac{4}{3}$

$\frac{1}{3}$

The a and b are the x-intercepts of the function.

The a and b are random x-values on the function.

The a and b are fixed points that the average value must always plug in, regardless of the interval of the function.

The a and b represent the endpoints of the interval over the function provided by the question.

Magnitude decreases as negative areas contribute less when calculating averages for real functions.

Magnitude increases due to absolute values being considered in averaging process.

No impact since multiplication by -1 doesn’t affect shape or area under curve within limits [-c,c].

The sign changes but magnitude stays same because integral values reflect area directionality (above/below x-axis).

Finding the average value is not likely to show up on an AP Calculus AB Exam.

MCQ only

FRQ only

Both MCQ and FRQ!

The average value becomes zero.

The average value decreases.

The average value increases.

The average value remains unchanged.

Its integral over [m,n] divided by (n-m) matches this average value.

The maximum value of h(t) occurs exactly at t = (m+n)/2.

It must have an inflection point somewhere within (m,n).

H(t)'s rate of change must be increasing throughout [m,n].

The function has positive values for some $x$ in the interval.

The function has negative values for some $x$ in the interval.

The function is always negative over the interval.

The function is always positive over the interval.